#### Answer

$z=\left\{ -3,3 \right\}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Express the given equation, $
9z^2-81=0
,$ in factored form. Then, use the Zero-Factor Property by equating each factor to zero. Finally, solve each equation.
$\bf{\text{Solution Details:}}$
The factored form of the equation above is
\begin{array}{l}\require{cancel}
9(z^2-9)=0
\\\\
9(z+3)(z-3)=0
\\\\
\dfrac{9(z+3)(z-3)}{9}=\dfrac{0}{9}
\\\\
(z+3)(z-3)=0
.\end{array}
Equating each factor to zero (Zero-Factor Property), then
\begin{array}{l}\require{cancel}
z+3=0
\text{ OR }
z-3=0
.\end{array}
Using the properties of equality to solve each of the equation above results to
\begin{array}{l}\require{cancel}
z+3=0
\\\\
z=-3
\\\\\text{ OR }\\\\
z-3=0
\\\\
z=3
.\end{array}
Hence, the solutions are $
z=\left\{ -3,3 \right\}
.$
$\bf{\text{Supplementary Solution/s:}}$
Factoring the $GCF=
9
,$ the expression, $
9z^2-81
,$ above is equivalent to
\begin{array}{l}\require{cancel}
9 \left( \dfrac{9z^2}{9}-\dfrac{81}{9} \right)
\\\\=
9 \left( z^2-9 \right)
.\end{array}
Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
9(z+3)(z-3)
.\end{array}